Show that $r^{2} = 3$ from a geometric series

Question

The sum of the first two terms of geometric sequence $a_n=a_0r^n$ is $7$, and the sum of the first six terms is $91$.

Show that $r^{2} = 3$.

How do we go about this?

Here's what I have tried -

$\displaystyle \Rightarrow \frac{\dfrac{a(r^{6}-1)}{r-1}}{\dfrac{a(r^{2}-1)}{r-1}} = \frac{91}{7}$

$\displaystyle \Rightarrow \frac{(r^{6}-1)}{(r^{2}-1)} = \frac{91}{7}$

$\Rightarrow\;???$

Any help would be highly appreciated.

Answer 1

Let $x=r^{2}$. The quation becomes $x^{3}-1=\frac {91} 7 (x-1)$ or $x^{2}+x+1=\frac {91} 7$ since $r=1$ is not valid. So $x=\frac {-1 \pm \sqrt {49}} 2$. Can you finish?

Answer 2

Use this identity $a^3- b^3 = (a-b)(a^2 +ab+b^2)$, and then solve a quadratic equation.

$$\Rightarrow \frac{(r^{6}-1)}{(r^{2}-1)} = 13$$ $$\Rightarrow \frac{(r^{2}-1)(r^4+r^2+1)}{(r^{2}-1)} = 13$$ $$\Rightarrow r^4+r^2+ 1= 13$$ $$\Rightarrow r^4+r^2-12=0$$

You'll find something like $(r^2 -3)(r^2 +4)=0$

BINGO!

Answer 3

Note that $$r^6-1=(r^3+1)(r^3-1).$$

From @MMM's answer, we can observe that for all $a$ and $b$, $$a^3-b^3=(a-b)(a^2+ab+b^2). $$ Let $a=r$ and $b=\pm 1$. $$\therefore r^6-1=(r+1)(r^2-r+1)\cdot (r-1)(r^2+r+1).$$ Since $r^2-1=(r+1)(r-1)$, it follows then that $$(r^2-r+1)(r^2+r+1)=13.$$

Can you take it from here? :)